why does n+1/n^2 converge and show simplification of the expression in detail

To understand why the series \(\sum_{n=1}^{\infty} \frac{n+1}{n^2}\) converges, we need to analyze and simplify the expression \(\frac{n+1}{n^2}\) and then examine its behavior as \(n\) becomes large.

### 1. **Simplifying the Expression**
Let's start by simplifying the expression \(\frac{n+1}{n^2}\):

\[
\frac{n+1}{n^2} = \frac{n}{n^2} + \frac{1}{n^2}
\]

Now, simplify each term individually:

\[
\frac{n}{n^2} = \frac{1}{n} \quad \text{and} \quad \frac{1}{n^2} \text{ is already simplified.}
\]

So, the expression becomes:

\[
\frac{n+1}{n^2} = \frac{1}{n} + \frac{1}{n^2}
\]

### 2. **Analyzing the Series**
The series \(\sum_{n=1}^{\infty} \frac{n+1}{n^2}\) can now be written as:

\[
\sum_{n=1}^{\infty} \left(\frac{1}{n} + \frac{1}{n^2}\right) = \sum_{n=1}^{\infty} \frac{1}{n} + \sum_{n=1}^{\infty} \frac{1}{n^2}
\]

This splits the original series into two separate series:

1. **The Harmonic Series** \(\sum_{n=1}^{\infty} \frac{1}{n}\):
   - This series is known to **diverge**.
   
2. **A p-Series** \(\sum_{n=1}^{\infty} \frac{1}{n^2}\):
   - This is a p-series with \(p = 2\), which is greater than 1. Such a series **converges**.

### 3. **Combined Behavior of the Series**
Even though the harmonic series diverges, the term \(\frac{1}{n^2}\) decreases much faster than \(\frac{1}{n}\) as \(n\) increases. The rapid decay of \(\frac{1}{n^2}\) is sufficient to make the combined series \(\sum_{n=1}^{\infty} \left(\frac{1}{n} + \frac{1}{n^2}\right)\) converge.

### 4. **Formal Convergence Argument**
To confirm convergence, consider the **limit comparison test**:

- We already know \(\sum \frac{1}{n^2}\) converges. Let's compare \(\frac{n+1}{n^2}\) with \(\frac{2}{n^2}\) as \(n\) grows large:
  
\[
\lim_{n \to \infty} \frac{\frac{n+1}{n^2}}{\frac{1}{n^2}} = \lim_{n \to \infty} \left(\frac{n+1}{1}\right) = \lim_{n \to \infty} \left(n + 1\right) = \infty
\]

Instead, let's compare it directly to the harmonic series, where:

\[
\frac{n+1}{n^2} \leq \frac{2n}{n^2} = \frac{2}{n}
\]

Thus, we have:

\[
\sum \frac{n+1}{n^2} \sim \sum \frac{2}{n^2}
\]

Since \(\sum \frac{2}{n^2}\) converges, the original series \(\sum \frac{n+1}{n^2}\) must also converge.

### 5. **Conclusion**
The series \(\sum_{n=1}^{\infty} \frac{n+1}{n^2}\) converges because, although it contains a divergent harmonic series term \(\frac{1}{n}\), the addition of the convergent term \(\frac{1}{n^2}\) ensures the overall series decreases rapidly enough to converge.

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Would you like more details or have any questions?

Here are some related questions:

1. What is the limit comparison test and how does it work for series convergence?
2. How does the behavior of \(\frac{1}{n}\) compare to \(\frac{1}{n^2}\) as \(n\) increases?
3. Why is the harmonic series divergent even though it decreases?
4. What are other examples of series that converge due to the rapid decrease of their terms?
5. How do we determine the convergence of a series involving a combination of divergent and convergent terms?

**Tip:** Simplifying the general term of a series and comparing it to a known convergent or divergent series is often a useful strategy for determining convergence.

Explore why the series \(\sum_{n=1}^{\infty} \frac{n+1}{n^2}\) converges. Detailed analysis includes simplifying \(\frac{n+1}{n^2}\), comparison with the harmonic series, and application of the limit comparison test. Suitable for advanced undergraduate level mathematics.

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